How many types of n-digit integers are there, if one is a permutation of another Two n-digit sequences of digits 0, 1.....9 are said to be of the same
type if the digits of one are a permutation of the digits of the other.
For n = 8, for example, the sequences 03088929 and 00238899 are the
same type. How many types of n-digit integers are there?
 A: This kind of problem can be usually be solved in various waysI always try to translate into counting lattice paths across a rectangle, since that gives easy comparison of solutions and less risk of off-by-one errors. The number of paths across a $a\times b$ rectangle is $\binom{a+b}a$, since the length of any path is $a+b$ steps, and we have to choose $a$ horizontal (say) steps among them, the remainder being vertical. Here are two possibilities.
The classes of numbers are characterised by their histogram of the digit frequencies. However that is ragged-shaped, so instead make the cumulative histogram, tabulating how many digits${}<1$, how many${}<2$, up to how many digits ${}<9$ (we know there are no digits${}<0$ and $n$ digits${}<10$, so there is no need to record those numbers). Since column lengths weakly increase, the boundary of the histogram is a path across a rectangle $9$ squares wide and $n$ squares high.
Another approach is to sort the digits into weakly increasing order, and represent each digit by a column of that height. Again the boundary is a path across a rectangle with left and up steps only; this time the rectangle is $n$ places wide and $9$ high.
Either way you find $\binom{n+9}9=\binom{n+9}n$ solutions.  
A: I would reformulate the problem as follows. For $k\in\{0,1,\dots,9\}$ let $x_k$ be the number of copies of the digit $k$ in your $n$-digit type; then 
$$x_0+x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9=n\;,\tag{1}$$ 
and each of the $x_k$’s is a non-negative integer. Conversely, each solution of $(1)$ in non-negative integers describes a type of $n$-digit number. The number of solutions in non-negative integers of $(1)$ is well-known; as Henning said in the comments, this is a stars-and-bars problem, and the solution is
$$\binom{n+10-1}{10-1}=\binom{n+9}9=\binom{n+9}n\;.$$
The reasoning behind this solution is explained quite well in the linked article.
